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283 Find the unit digit of $\left(\frac{5+\sqrt{21}}{2}\right)^{2010}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve the problem, which is to find the remainder when $\left[\left(\frac{5+\sqrt{21}}{2}\right)^{2010}\right]$ is divided by 10.
Let $a_{n}=\left(\frac{5+\sqrt{21}}{2}\right)^{n}+\left(\frac{5-\sqrt{21}}{2}\right)^{n}\left(n \in \mathbf{N}_{+}\right)$.
Since $\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=5$,
$$
\frac{5+... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 On a plane, there are seven points, and some line segments can be connected between them, so that any three points among the seven must have at least one line segment connecting a pair of them. How many line segments are needed at least? Prove your conclusion.
$(2002$, Shanghai Junior High School Mathematics ... | (1) If point $A$ is not an endpoint of any line segment, then every two of the other six points are connected by a line segment, totaling $\frac{1}{2} \times 6 \times 5=15$ line segments;
(2) If point $A$ is the endpoint of only one line segment, then every two of the five points not connected to $A$ are connected by a... | 9 | Combinatorics | proof | Yes | Yes | cn_contest | false |
8. Given 8 points $A_{1}, A_{2}, \cdots, A_{8}$ on a circle. Find the smallest positive integer $n$, such that among any $n$ triangles with these 8 points as vertices, there must be two triangles that share a common side.
(Tao Pingsheng, problem contributor) | 8. First, consider the maximum number of triangles with no common edges.
Connecting every pair of eight points yields $\mathrm{C}_{8}^{2}=28$ chords.
If each chord belongs to only one triangle, then these chords can form at most $r \leqslant\left[\frac{28}{3}\right]=9$ triangles with no common edges. However, if ther... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Robots A and B start from the starting point at the same time, moving uniformly along a hundred-meter track, and the automatic recorder shows: when A is $1 \mathrm{~m}$ away from the finish line, B is $2 \mathrm{~m}$ away from the finish line; when A reaches the finish line, B is $1.01 \mathrm{~m}$ away from the fin... | 2.1.
Let the actual length of the track be $x \mathrm{~m}$, and the speeds of robots 甲 and 乙 be $v_{\text {甲 }}$ and $v_{\text {乙}}$, respectively. Thus,
$$
\frac{v_{\text {甲 }}}{v_{\text {乙 }}}=\frac{x-1}{x-2}=\frac{x}{x-1.01} \text {. }
$$
Solving for $x$ yields $x=101$.
Therefore, this track is $1 \mathrm{~m}$ lon... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Among the seven points consisting of the center and the six vertices of a regular hexagon, if any $n$ points are taken, and among them, there must be three points that form the vertices of an equilateral triangle, then the minimum value of $n$ is $\qquad$ | -1.5 .
Consider the regular hexagon $A B C D E F$ as shown in Figure 1, with its center at $O$.
When $n=4$, take $A, C, D, F$, among which no three points can form the three vertices of an equilateral triangle.
When $n=5$, consider the following two cases:
(1) $O$ is among the five points. Consider the three pairs of ... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. As shown in Figure 1, in the cube $A C_{1}$ with edge length 1, points $P$ and $Q$ are moving points on edges $A D$ and $A_{1} B_{1}$, respectively. If the skew lines $B D_{1}$ and $P Q$ are perpendicular to each other, then $A P + A_{1} Q =$ $\qquad$ | 5. 1 .
As shown in Figure 5, establish a spatial rectangular coordinate system, and set
$$
\begin{array}{l}
B(1,0,0), \\
D_{1}(0,1,1), \\
P(0, a, 0), \\
Q(b, 0,1) . \\
\text { Then } \overrightarrow{B D_{1}}=(-1,1,1), \\
\overrightarrow{P Q}=(b,-a, 1) .
\end{array}
$$
$$
\text { Therefore, } 0=\overrightarrow{B D_{1}... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. For a positive integer $x$, let $S(x)$ denote the sum of the digits of $x$. Then the maximum value of $S(x)-9[\lg x]$ is $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 8. 9 .
Let $x=\overline{a_{n} a_{n-1} \cdots a_{0}}\left(a_{0} \neq 0\right)$. Then
$$
\begin{array}{l}
10^{n} \leqslant x<10^{n+1} \\
\Rightarrow n \leqslant \lg x<n+1 \Rightarrow[\lg x]=n . \\
\text { Hence } S(x)=a_{0}+a_{1}+\cdots+a_{n} \leqslant 9(n+1) \\
=9([\lg x]+1) .
\end{array}
$$
Therefore, $S(x)-9[\lg x] ... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let $n$ be a positive integer, and $f(n)$ denote the number of integers satisfying the following conditions:
(i) Each digit $a_{i} \in\{1,2,3,4\}$, and
$a_{i} \neq a_{i+1}(i=1,2, \cdots)$;
(ii) When $n \geqslant 3$, $a_{i}-a_{i+1}$ and $a_{i+1}-a_{i+2}$ $(i=1,2, \cdots)$ have opposite signs.
(1) Find the valu... | Let $g(n)$ denote the number of wave numbers for which $a_{n}>a_{n-1}$ when $n \geqslant 2$. Then, by symmetry, we have
$$
\begin{array}{l}
g(n)=\frac{1}{2} f(n) . \\
\text { Hence } a_{n-1}=1, a_{n}=2,3,4 ; \\
a_{n-1}=2, a_{n}=3,4 ; \\
a_{n-1}=3, a_{n}=4 .
\end{array}
$$
Let $m(i)$ represent the number of $(n-1)$-dig... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $f(x)$ represent a fourth-degree polynomial in $x$. If
$$
f(1)=f(2)=f(3)=0, f(4)=6, f(5)=72 \text {, }
$$
then the last digit of $f(2010)$ is $\qquad$ | 6. 2 .
Since $f(1)=f(2)=f(3)=0$, the quartic polynomial can be set as
$$
\begin{array}{l}
f(x)=(x-1)(x-2)(x-3)(a x+b) . \\
\text { By } f(4)=6, f(5)=72, \text { we get } \\
\left\{\begin{array} { l }
{ 6 ( 4 a + b ) = 6 , } \\
{ 2 4 ( 5 a + b ) = 7 2 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=2, \\
b=-7 .
\e... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $p, q$ be prime numbers, and satisfy $p^{3}+q^{3}+1=p^{2} q^{2}$. Then the maximum value of $p+q$ is | 8.5.
Assume $q \leqslant p$. Notice that
$$
\begin{array}{l}
p^{3}+q^{3}+1=p^{2} q^{2} \\
\Rightarrow q^{3}+1=p^{2} q^{2}-p^{3} \\
\Rightarrow(q+1)\left(q^{2}-q+1\right)=p^{2}\left(q^{2}-p\right) .
\end{array}
$$
Therefore, $p^{2} \mid(q+1)\left(q^{2}-q+1\right)$.
Since $q \leqslant p$, we have $0<q^{2}-q+1<p^{2}$.
T... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Line segment $A B$ divides a square into two polygons (points $A$ and $B$ are on the sides of the square), each polygon has an inscribed circle, one of which has a radius of 6, while the other has a radius greater than 6. What is the difference between the side length of the square and twice the length of line segme... | If line segment $A B$ divides the square into two triangles, then $A B$ can only be the diagonal of the square. But in this case, the radii of the two incircles are equal, which contradicts the condition.
If one of the polygons is a quadrilateral, then the sum of the lengths of $A B$ and its opposite side is greater t... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. (40 points) Let integers $m, n$ satisfy $m \geqslant n$ and $m^{3}+n^{3}+1=4 m n$. Find the maximum value of $m-n$.
保留了原文的换行和格式，如上所示。 | 3. Let $s=m+n, p=mn$. Then $m^{3}+n^{3}=(m+n)^{3}-3m^{2}n-3mn^{2}$ $=s^{3}-3ps$.
From $s^{3}-3ps+1=4p$, we get $p=\frac{s^{3}+1}{3s+4}$.
Since $m, n$ are integers, 27p must also be an integer, and
$27p=9s^{2}-12s+16-\frac{37}{3s+4}$.
Therefore, $3s+4$ must divide 37.
Thus, $3s+4 = \pm 1$ or $\pm 37$.
If $3s+4=-1$ or -3... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. (40 points) Write "KOREAIMC" in eight lines as shown in Figure 5, where the first line has one K, the second line has two O's, and so on, with the last line having eight C's.
$$
\begin{array}{llllllll}
\mathbf{K} & & & & & & & \\
\mathbf{O} & \mathbf{O} & & & & & & \\
\mathbf{R} & \mathbf{R} & \mathbf{R} & & & & & ... | 10. First list the number table as shown in Figure 17, where each cell corresponds to a letter, and the number in the cell represents the number of different paths from the letter "K" to that letter. The cell at the very top is filled with 1. Starting from the second row, the number in each cell is the sum of the numbe... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $k$ be an integer greater than 1, and the sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
\begin{array}{l}
a_{0}=0, a_{1}=1, \\
a_{n+1}=k a_{n}+a_{n-1}(n=1,2, \cdots) .
\end{array}
$$
Find all $k$ that satisfy the following condition: there exist non-negative integers $l, m (l \neq m)$, and positive i... | When $k=2$, $a_{0}=0, a_{1}=1, a_{2}=2$, then from $a_{0}+2 a_{2}=a_{2}+2 a_{1}=4$, we know that taking $l=0, m=2, p=$ $2, q=1$ is sufficient.
For $k \geqslant 3$, by the recurrence relation, $\left\{a_{n}\right\}$ is a strictly increasing sequence of natural numbers and $k \mid\left(a_{n+1}-a_{n-1}\right)$.
$$
\begin... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Find the value of $\frac{\log _{2} \frac{2}{3}+\log _{2} \frac{3}{4}+\log _{2} \frac{4}{5}+\log _{2} \frac{5}{6}+\log _{2} \frac{6}{7}+\log _{2} \frac{7}{8}}{\log _{3} 3 \cdot \log _{4} 3 \cdot \log _{5} 4 \cdot \log _{8} 5 \cdot \log _{7} 6 \cdot \log _{8} 7}$. | 2. -6 .
$$
\begin{array}{l}
\text { Original expression }=\frac{\log _{2}\left(\frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \cdot \frac{6}{7} \cdot \frac{7}{8}\right)}{\frac{\lg 2}{\lg 3} \cdot \frac{\lg 3}{\lg 4} \cdot \frac{\lg 4}{\lg 5} \cdot \frac{\lg 5}{\lg 6} \cdot \frac{\lg 6}{\lg 7} \cdot \... | -6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $x_{1}, x_{2}, \cdots, x_{2010}$ are all positive real numbers. Then
$$
x_{1}+\frac{x_{2}}{x_{1}}+\frac{x_{3}}{x_{1} x_{2}}+\cdots+\frac{x_{2010}}{x_{1} x_{2} \cdots x_{200}}+\frac{4}{x_{1} x_{2} \cdots x_{2010}}
$$
the minimum value is $\qquad$ | 3.4.
Starting from the last two terms, repeatedly applying the AM-GM inequality, we get
$$
\begin{array}{l}
\text { Original expression }=\sum_{i=1}^{2010} \frac{x_{i}}{\prod_{j=1}^{i-1} x_{j}}+\frac{4}{\prod_{j=1}^{2010} x_{j}} \\
=\sum_{i=1}^{2009} \frac{x_{i}}{\prod_{j=1}^{i-1} x_{j}}+\left(\frac{x_{2010}}{\prod_{j... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that the circumcenter, incenter, and orthocenter of a non-isosceles acute $\triangle ABC$ are $O, I, H$ respectively, and $\angle A=60^{\circ}$. If the altitudes of $\triangle ABC$ are $AD, BE, CF$, then the ratio of the circumradius of $\triangle OIH$ to the circumradius of $\triangle DEF$ is $\qquad$ . | 4. 2 .
From $\angle B O C=\angle B I C=\angle B H C=120^{\circ}$, we know that $O, I, H, B, C$ are concyclic.
Let the circumradii of $\triangle A B C$ and $\triangle O I H$ be $R$ and $r$, respectively.
By the Law of Sines, we have
$2 R \sin A=B C=2 r \sin \angle B O C$.
Thus, $r=R$.
Let the circumradius of $\triangl... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. (15 points) In a regular pentagon $A B C D E$, the diagonal $B E$ intersects diagonals $A D$ and $A C$ at points $F$ and $G$, respectively. The diagonal $B D$ intersects diagonals $C A$ and $C E$ at points $H$ and $I$, respectively. The diagonal $C E$ intersects diagonal $A D$ at point $J$. Let the set of isosceles... | 11. (1) Since all triangles formed by 10 points $A, B, C, D, E, F, G, H, I, J$ and line segments in Figure 2 are isosceles triangles,
$$
|M|=\mathrm{C}_{5}^{3}+4 \mathrm{C}_{5}^{4}+5 \mathrm{C}_{5}^{5}=35 \text{. }
$$
(2) By the pigeonhole principle, we know that among $A, B, C, D, E$, there must be three points of the... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Divide a wire of length $11 \mathrm{~cm}$ into several segments of integer centimeters, such that any three segments can form the sides of a triangle. Then the number of different ways to divide the wire is $\qquad$ (ways with the same number of segments and corresponding equal lengths are considered the same way). | 4.8.
Let the shortest two segments be $a \mathrm{~cm}, b \mathrm{~cm}$, and take any other segment $c \mathrm{~cm} (a \leqslant b \leqslant c)$. Then $b \leqslant c < a + b$.
(1) If $a = 1$, then $1 \leqslant b \leqslant c < b + 1$. Thus, $b = c$. Therefore, $b \in (11 - 1), 2b < 11 - 1$. Hence, $b$ can be $1, 2, 5$, ... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
14. In space, there are five points, no four of which are coplanar. If several line segments are drawn such that no tetrahedron exists in the graph, then the maximum number of triangles in the graph is $\qquad$. | 14.4.
First, construct graph 6. It is easy to see that it meets the conditions and has exactly four triangles.
Now assume there exists some configuration where the number of triangles is no less than five.
If only two line segments are not connected, then these two line segments must have no common endpoints (as sho... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
18. (17 points) Let $a_{1}, a_{2}, \cdots, a_{n}$ be a permutation of the integers 1, 2, $\cdots, n$, and satisfy the following conditions:
(1) $a_{1}=1$;
(2) $\left|a_{i}-a_{i+1}\right| \leqslant 2(i=1,2, \cdots, n-1)$.
Let the number of such permutations be $f(n)$. Find the remainder when $f(2010)$ is divided by 3. | 18. Verifiable
$$
f(1)=1, f(2)=1, f(3)=2 \text {. }
$$
Let $n \geqslant 4$. Then $a_{2}=2$ or 3.
For $a_{2}=2$, the number of permutations is $f(n-1)$. This is because by removing the first term and reducing all subsequent terms by 1, a one-to-one correspondence can be established.
For $a_{2}=3$, if $a_{3}=2$, then $... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Find the smallest positive integer $n$, such that among any $n$ irrational numbers, there are always three numbers, the sum of any two of which is irrational. | Three, take four irrational numbers $\{\sqrt{2}, \sqrt{3},-\sqrt{2},-\sqrt{3}\}$, obviously they do not satisfy the condition, hence $n \geqslant 5$.
Consider five irrational numbers $a, b, c, d, e$, viewed as five points.
If the sum of two numbers is a rational number, then connect the corresponding two points with a ... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. From the set $\{1,2, \cdots, 2011\}$, randomly select 1005 different numbers such that their sum is 1021035. Then, there are at least some odd numbers among them. | 6.5.
From 1 to 2011, taking 1005 even numbers, their sum is $\frac{1005(2+2010)}{2}=1011030$.
Replacing $2,4,6,8,10$ with 2003, 2005, $2007,2009,2011$, increases the sum by
$$
2001 \times 5=10005,
$$
making the total sum 1021035.
Thus, there are at least 5 odd numbers. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Inside a large sphere with a radius of 4, 24 cubes with edge lengths of 1 have been placed arbitrarily. Prove: At least 4 small spheres with a radius of $\frac{1}{2}$ can still be placed inside the large sphere, such that these small spheres and the cubes are all within the large sphere and do not overl... | 11. To place a small ball with a radius of $\frac{1}{2}$ completely inside the large ball $D(O, 4)$, the distance from its center to the surface of the large ball should be no less than $\frac{1}{2}$, meaning the center of the small ball should be within the ball $D\left(0,4-\frac{1}{2}\right)$. The volume of this ball... | 4 | Geometry | proof | Yes | Yes | cn_contest | false |
For any positive integer $n(n \geqslant 2)$, try to find:
$$
\begin{array}{l}
\sum_{k=2}^{n}\left[\log _{\frac{3}{2}}\left(k^{3}+1\right)-\log _{\frac{3}{2}}\left(k^{3}-1\right)\right]+ \\
\frac{1}{2} \log _{\frac{3}{2}}\left[1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}\right]
\end{array}
$$
the value. | Notice
$$
\begin{array}{l}
\sum_{k=2}^{n}\left[\log _{\frac{3}{2}}\left(k^{3}+1\right)-\log _{\frac{3}{2}}\left(k^{3}-1\right)\right] \\
=\sum_{k=2}^{n} \log _{\frac{3}{2}} \frac{k^{3}+1}{k^{3}-1}=\log _{\frac{3}{2}} \prod_{k=2}^{n} \frac{k^{3}+1}{k^{3}-1} \\
=\log _{\frac{3}{2}} \prod_{k=2}^{n} \frac{(k+1)\left(k^{2}-... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Given the quadratic function
$$
y=a x^{2}+b x+c \geqslant 0(a<b) \text {. }
$$
Then the minimum value of $M=\frac{a+2 b+4 c}{b-a}$ is $\qquad$ | 11. 8.
From the conditions, it is easy to see that $a>0, b^{2}-4 a c \leqslant 0$.
Notice that
$$
\begin{array}{l}
M=\frac{a+2 b+4 c}{b-a}=\frac{a^{2}+2 a b+4 a c}{a(b-a)} \\
\geqslant \frac{a^{2}+2 a b+b^{2}}{a(b-a)} .
\end{array}
$$
Let $t=\frac{b}{a}$. Then $t>1$. Thus,
$$
\begin{array}{l}
M \geqslant \frac{a^{2}+2... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
16. (15 points) As shown in Figure 2, it is known that the ellipse $C$ passes through the point $M(2,1)$, with the two foci at $(-\sqrt{6}, 0)$ and $(\sqrt{6}, 0)$. $O$ is the origin, and a line $l$ parallel to $OM$ intersects the ellipse $C$ at two different points $A$ and $B$.
(1) Find the maximum value of the area o... | 16. (1) Let the equation of the ellipse $C$ be
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0) \text {. }
$$
From the problem, we have
$$
\left\{\begin{array} { l }
{ a ^ { 2 } - b ^ { 2 } = 6 , } \\
{ \frac { 4 } { a ^ { 2 } } + \frac { 1 } { b ^ { 2 } } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a^{2... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
17. (15 points) Given the function
$$
f(x)=\frac{1}{2} m x^{2}-2 x+1+\ln (x+1)(m \geqslant 1) \text {. }
$$
(1) If the curve $C: y=f(x)$ has a tangent line $l$ at point $P(0,1)$ that intersects $C$ at only one point, find the value of $m$;
(2) Prove that the function $f(x)$ has a decreasing interval $[a, b]$, and find ... | 17. (1) Note that the domain of the function $f(x)$ is $(-1,+\infty)$,
$$
f^{\prime}(x)=m x-2+\frac{1}{x+1}, f^{\prime}(0)=-1 .
$$
Therefore, the slope of the tangent line $l$ at the point of tangency $P(0,1)$ is -1.
Thus, the equation of the tangent line is $y=-x+1$.
Since the tangent line $l$ intersects the curve $C... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let real numbers $s, t$ satisfy
$$
\begin{array}{l}
19 s^{2}+99 s+1=0, \\
t^{2}+99 t+19=0(s t \neq 1) . \\
\text { Find the value of } \frac{s t+4 s+1}{t} \text { . }
\end{array}
$$
(1999, National Junior High School Mathematics Competition) | 【Analysis】Transform the first equation of the known equations, and combine it with the second equation to find that $\frac{1}{s} 、 t(s t \neq 1)$ are the two roots of the quadratic equation
$$
x^{2}+99 x+19=0
$$
Since $s \neq 0$, the first equation can be transformed into
$$
\left(\frac{1}{s}\right)^{2}+99\left(\frac{... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x+y=z-1, \\
x y=z^{2}-7 z+14 .
\end{array}\right.
$$
Question: What is the maximum value of $x^{2}+y^{2}$? For what value of $z$ does $x^{2}+y^{2}$ achieve its maximum value? | Prompt: Example 6. From the problem, we know that $x$ and $y$ are the two real roots of the equation
$$
t^{2}-(z-1) t+z^{2}-7 z+14=0
$$
By the discriminant $\Delta \geqslant 0$, we get
$$
3 z^{2}-26 z+55 \leqslant 0 \text {. }
$$
Solving this, we get $\frac{11}{3} \leqslant z \leqslant 5$.
$$
\begin{array}{l}
\text {... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
the integer part.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Let
then $2<\sqrt[3]{23}=a_{1}<3, a_{n+1}=\sqrt[3]{23+a_{n}}$.
By mathematical induction, it is easy to prove
$$
2<a_{n}<3\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
In fact,
the integer part is 2. | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given real numbers $x, y$ satisfy
$$
x^{2}+3 y^{2}-12 y+12=0 \text {. }
$$
then the value of $y^{x}$ is $\qquad$ | Hint: Treat the known equation as a quadratic equation in $y$ (the main variable)
$$
3 y^{2}-12 y+\left(12+x^{2}\right)=0 \text {. }
$$
From $\Delta=-12 x^{2} \geqslant 0$, and since $x^{2} \geqslant 0$, then $x=0$.
Thus, $y=2$. Therefore, $y^{x}=2^{0}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given real numbers $a, b, c$ satisfy
$$
a=2 b+\sqrt{2}, a b+\frac{\sqrt{3}}{2} c^{2}+\frac{1}{4}=0 \text {. }
$$
Then $\frac{b c}{a}=$ $\qquad$ . | Hint: Eliminate $a$ from the two known equations, then treat $c$ as a constant, to obtain a quadratic equation in $b$
$$
2 b^{2}+\sqrt{2} b+\frac{\sqrt{3}}{2} c^{2}+\frac{1}{4}=0 .
$$
From the discriminant $\Delta \geqslant 0$, we get $c=0$.
Therefore, $\frac{b c}{a}=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given
$\frac{1}{4}(b-c)^{2}=(a-b)(c-a)$, and $a \neq 0$.
Then $\frac{b+c}{a}=$ $\qquad$ | From the given equation, we have
$$
(b-c)^{2}-4(a-b)(c-a)=0 \text {. }
$$
When $a \neq b$, by the discriminant of a quadratic equation, we know that the quadratic equation in $x$
$$
(a-b) x^{2}+(b-c) x+(c-a)=0
$$
has two equal real roots.
$$
\text { Also, }(a-b)+(b-c)+(c-a)=0 \text {, thus, }
$$
$x=1$ is a root of eq... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (15 points) Can 2010 be written as the sum of squares of $k$ distinct prime numbers? If so, try to find the maximum value of $k$; if not, please briefly explain the reason. | If 2010 can be written as the sum of squares of $k$ prime numbers, taking the sum of the squares of the smallest 10 distinct primes, then
$$
\begin{array}{l}
4+9+25+49+121+169+ \\
289+361+529+841 \\
=2397>2010 .
\end{array}
$$
Therefore, $k \leqslant 9$.
It is easy to see that there is only one even prime number 2, an... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given points $M(0,2)$ and $N(-3,6)$, the distances from these points to line $l$ are $1$ and $4$, respectively. The number of lines $l$ that satisfy these conditions is . $\qquad$ | 2.3.
It is easy to get $M N=5$.
Then the circle $\odot M$ with radius 1 is externally tangent to the circle $\odot N$ with radius 4. Therefore, there are 3 lines $l$ that satisfy the condition. | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=2, a_{n+1}=\frac{1+a_{n}}{1-a_{n}}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Let $T_{n}=a_{1} a_{2} \cdots a_{n}$. Then $T_{2010}=$ $\qquad$ | 5. -6 .
It is easy to get $a_{1}=2, a_{2}=-3, a_{3}=-\frac{1}{2}, a_{4}=\frac{1}{3}$, $a_{1} a_{2} a_{3} a_{4}=1$.
Also, $a_{5}=2=a_{1}$, by induction it is easy to know $a_{n+4}=a_{n}\left(n \in \mathbf{N}_{+}\right)$.
Therefore, $T_{2010}=T_{4 \times 502+2}=a_{1} a_{2}=-6$. | -6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $a=\frac{\sqrt{5}-1}{2}$. Then $\frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a}=$ $\qquad$ | Given $a=\frac{-1+\sqrt{5}}{2}$, we can obtain the corresponding quadratic equation $a^{2}+a-1=0$.
Thus, the original expression is
$$
\begin{array}{l}
=\frac{\left(a^{2}+a-1\right)\left(a^{3}-a\right)-2(a-1)}{a\left(a^{2}-1\right)} \\
=\frac{-2}{a^{2}+a}=-2 .
\end{array}
$$ | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the sequence $\left\{x_{n}\right\}$ :
$$
1,3,3,3,5,5,5,5,5, \cdots
$$
formed by all positive odd numbers arranged from smallest to largest, and each odd number $k(k=1,3,5, \cdots)$ appears consecutively $k$ times. If the general term formula of this sequence is $x_{n}=a[\sqrt{b n+c}]+d$, then $a+b+c+d=$ $\qqu... | 4.3.
For $k^{2}+1 \leqslant n \leqslant(k+1)^{2}$, $x_{n}=2 k+1, k=[\sqrt{n-1}]$,
therefore, $x_{n}=2[\sqrt{n-1}]+1$.
Thus, $(a, b, c, d)=(2,1,-1,1)$.
So $a+b+c+d=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. $\sum_{i=0}^{50} \sum_{j=0}^{50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}$ modulo 31 is
$\qquad$ . | 6.1.
Original expression $=\sum_{i=0}^{50} C_{50}^{i} \cdot \sum_{j=0}^{50} C_{50}^{j}=\left(2^{50}\right)^{2}=2^{100}$.
And $2^{5} \equiv 1(\bmod 31)$, so
Original expression $=2^{100} \equiv 1(\bmod 31)$.
Therefore, the remainder is 1. | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. In a certain kingdom, there are 32 knights, some of whom are servants to other knights. Each servant can have at most one master, and each master must be richer than any of his servants. If a knight has at least four servants, he is ennobled as a noble. If it is stipulated that a servant of $A$'s servant is not a se... | 2.7.
According to the problem, the richest knight is not a servant of any other knight, so at most 31 knights can be servants of other knights.
Since each noble has at least four servants, there can be at most 7 nobles.
Number the 32 knights as $1,2, \cdots, 32$, with wealth decreasing as the number increases, then
... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
II. (25 points) Given 15 quadratic equations $x^{2}-p_{i} x+q_{i}=0(i=1,2, \cdots, 15)$ with coefficients $p_{i} 、 q_{i}$ taking values from $1,2, \cdots, 30$, and these coefficients are all distinct. If an equation has a root greater than 20, it is called a "good equation." Find the maximum number of good equations. | Second, if there exists an equation $x^{2}-p_{i} x+q_{i}=0$ with two roots $x_{1}, x_{2}$, then from $x_{1}+x_{2}=p_{i}>0, x_{1} x_{2}=q_{i}>0$, we know that both roots are positive.
Next, if a certain equation has a root greater than 20, then $p_{i}=x_{1}+x_{2}>20$. However, among the numbers $1,2, \cdots, 30$, there... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) As shown in Figure 2, given points $A$ and $B$ are two distinct points outside circle $\odot O$, point $P$ is on $\odot O$, and $PA$, $PB$ intersect $\odot O$ at points $D$ and $C$ respectively, different from point $P$, and $AD \cdot AP = BC \cdot BP$.
(1) Prove: $\triangle OAB$ is an isosceles tria... | (1) Draw $A T$ tangent to $\odot O$ at point $T$, and connect $O T$.
Let the radius of $\odot O$ be $R$. Then
$$
A D \cdot A P=A T^{2}=O A^{2}-R^{2} \text {. }
$$
Similarly, $B C \cdot B P=O B^{2}-R^{2}$.
From the given, it is easy to see that $O A=O B$.
(2) From the proof in (1), we know
$$
p(2 p+1)=(m-1)^{2}-9 \text... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the inequality
$$
3 x+4 \sqrt{x y} \leqslant a(x+y)
$$
holds for all positive numbers $x$ and $y$. Then the minimum value of the real number $a$ is . $\qquad$ | -1.4 .
From the problem, we know that $a \geqslant\left(\frac{3 x+4 \sqrt{x y}}{x+y}\right)_{\text {max }}$.
$$
\text { Also, } \frac{3 x+4 \sqrt{x y}}{x+y} \leqslant \frac{3 x+(x+4 y)}{x+y}=4 \text {, }
$$
with equality holding if and only if $x=4 y>0$.
Therefore, the minimum value of $a$ is 4. | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. $[x]$ is the greatest integer not exceeding the real number $x$. It is known that the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{3}{2}, a_{n+1}=a_{n}^{2}-a_{n}+1\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then $m=\left[\sum_{k=1}^{2011} \frac{1}{a_{k}}\right]$ is . $\qquad$ | 4.1.
$$
\begin{array}{l}
\text { Given } a_{n+1}=a_{n}^{2}-a_{n}+1 \\
\Rightarrow a_{n+1}-1=a_{n}\left(a_{n}-1\right) \\
\Rightarrow \frac{1}{a_{n+1}-1}=\frac{1}{a_{n}-1}-\frac{1}{a_{n}} . \\
\text { Then } \frac{1}{a_{n}}=\frac{1}{a_{n}-1}-\frac{1}{a_{n+1}-1} . \\
\text { Therefore } m=\left[\frac{1}{a_{1}-1}-\frac{1}... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and
$$
f(x)=f(1-x) \text {. }
$$
Then $f(2010)=$ $\qquad$ . | $$
\begin{array}{l}
f(0)=0, \\
f(x+1)=f(-x)=-f(x) .
\end{array}
$$
From the conditions, we have $f(x+2)=f(x)$, which means $f(x)$ is a periodic function with a period of 2.
Therefore, $f(2010)=f(0)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Given real numbers $x, y$ satisfy
$$
3|x+1|+2|y-1| \leqslant 6 \text {. }
$$
Then the maximum value of $2 x-3 y$ is $\qquad$ . | 10.4.
The figure determined by $3|x+1|+2|y-1| \leqslant 6$ is the quadrilateral $ABCD$ and its interior, where,
$$
A(-1,4) 、 B(1,1) 、 C(-1,-2) 、 D(-3,1) \text {. }
$$
By the knowledge of linear programming, the maximum value of $2x-3y$ is 4.
The maximum value can be achieved when $x=-1, y=-2$. | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the sequence $\left\{8 \times\left(-\frac{1}{3}\right)^{n-1}\right\}$ have the sum of its first $n$ terms as $S_{n}$. Then the smallest integer $n$ that satisfies the inequality
$$
\left|S_{n}-6\right|<\frac{1}{125}
$$
is | 2. 7 | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. For $0<x<1$, if the complex number
$$
z=\sqrt{x}+\mathrm{i} \sqrt{\sin x}
$$
corresponds to a point, then the number of such points inside the unit circle is $n=$ | 6. 1 .
From the point on the unit circle, we have
$$
x+\sin x=1(00(x \in(0,1))$, which means $\varphi(x)$ is a strictly increasing function.
Also, $\varphi(0)=-10$, so the equation $x+\sin x=1$ has exactly one real root in $(0,1)$.
Therefore, $n=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $\frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{2010}$.
Then $m=$ $\qquad$ $(\bmod 2011)$. | 8. 0 .
Notice
$$
\begin{array}{l}
\frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{2010} \\
=\left(\frac{1}{1}+\frac{1}{2010}\right)+\left(\frac{1}{2}+\frac{1}{2009}\right)+\cdots+ \\
\left(\frac{1}{1005}+\frac{1}{1006}\right) \\
= \frac{2011}{1 \times 2010}+\frac{2011}{2 \times 2009}+\cdots+\frac{2011}{1005 \times 1006} \\... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Consider the matrix
$$
\left(a_{i j}\right)_{n \times n}\left(a_{i j} \in\{1,2,3\}\right) \text {. }
$$
If $a_{i j}$ is such that its row $i$ and column $j$ both contain at least three elements (including $a_{i j}$) that are equal to $a_{i j}$, then the element $a_{i j}$ is called "good". If the matrix $\left(a_{i j}\... | The minimum value of $n$ is 7.
When $n=6$, the matrix
$\left(\begin{array}{llllll}1 & 1 & 2 & 2 & 3 & 3 \\ 1 & 1 & 3 & 3 & 2 & 2 \\ 2 & 2 & 1 & 1 & 3 & 3 \\ 2 & 2 & 3 & 3 & 1 & 1 \\ 3 & 3 & 2 & 2 & 1 & 1 \\ 3 & 3 & 1 & 1 & 2 & 2\end{array}\right)$
has no good elements, so $n \geqslant 7$.
Below, we use proof by contrad... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. B. Given that $\angle A, \angle B$ are two acute angles, and satisfy
$$
\begin{array}{l}
\sin ^{2} A+\cos ^{2} B=\frac{5}{4} t, \\
\cos ^{2} A+\sin ^{2} B=\frac{3}{4} t^{2} .
\end{array}
$$
Then the sum of all possible real values of $t$ is ().
(A) $-\frac{8}{3}$
(B) $-\frac{5}{3}$
(C) 1
(D) $\frac{11}{3}$ | 3. B. C.
Adding the two equations yields $3 t^{2}+5 t=8$.
Solving gives $t=1, t=-\frac{8}{3}$ (discard).
When $t=1$, $\angle A=45^{\circ}, \angle B=30^{\circ}$ satisfies the given equations.
Therefore, the sum of all possible values of the real number $t$ is 1. | 1 | Algebra | MCQ | Yes | Yes | cn_contest | false |
5. A. Let $S=\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{99^{3}}$. Then the integer part of $4 S$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | 5. A. A.
When $k=2,3, \cdots, 99$, we have
$$
\frac{1}{k^{3}}<\frac{1}{k\left(k^{2}-1\right)}=\frac{1}{2}\left[\frac{1}{(k-1) k}-\frac{1}{k(k+1)}\right] \text {, }
$$
Thus, $1<S=1+\sum_{k=2}^{29} \frac{1}{k^{3}}$
$$
<1+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99 \times 100}\right)<\frac{5}{4} \text {. }
$$
Therefore, $... | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false |
8. A. As shown in Figure 1, points $A$ and $B$ lie on the line $y=x$. Parallel lines to the $y$-axis through $A$ and $B$ intersect the hyperbola $y=\frac{1}{x} (x>0)$ at points $C$ and $D$. If $BD=2AC$, then the value of $4OC^2-OD^2$ is $\qquad$ | 8. A. 6 .
Let points $C(a, b), D(c, d)$.
Then points $A(a, a), B(c, c)$.
Since points $C, D$ are on the hyperbola $y=\frac{1}{x}$, we have $a b=1, c d=1$.
Given $B D=2 A C$
$$
\begin{array}{l}
\Rightarrow|c-d|=2|a-b| \\
\Rightarrow c^{2}-2 c d+d^{2}=4\left(a^{2}-2 a b+b^{2}\right) \\
\Rightarrow 4\left(a^{2}+b^{2}\rig... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. B. Let the four-digit number
$\overline{a b c d}$ satisfy
$$
a^{3}+b^{3}+c^{3}+d^{3}+1=10 c+d .
$$
Then the number of such four-digit numbers is $\qquad$ | 10. B. 5 .
From $d^{3} \geqslant d$ we know $c^{3}+1 \leqslant 10 c \Rightarrow 1 \leqslant c \leqslant 3$.
If $c=3$, then $a^{3}+b^{3}+d^{3}=2+d$.
Thus, $d=1$ or 0. Therefore, $a=b=1$, which gives us $1131, 1130$ as solutions.
If $c=2$, then $a^{3}+b^{3}+d^{3}=11+d$. Thus, $d \leqslant 2$.
When $d=2$, $a^{3}+b^{3}=5$... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $f(x)$ be a polynomial with integer coefficients, $f(0)=11$, and there exist $n$ distinct integers $x_{1}, x_{2}, \cdots, x_{n}$, such that
$$
f\left(x_{1}\right)=f\left(x_{2}\right)=\cdots=f\left(x_{n}\right)=2010 .
$$
Then the maximum value of $n$ is | 7.3.
Let $g(x)=f(x)-2010$. Then $x_{1}, x_{2}, \cdots, x_{n}$ are all roots of $g(x)=0$.
Thus, $g(x)=\prod_{i=1}^{n}\left(x-x_{i}\right) \cdot q(x)$,
where $q(x)$ is a polynomial with integer coefficients. Therefore,
$$
\begin{array}{l}
g(0)=11-2010=-1999 \\
=\prod_{i=1}^{n}\left(-x_{i}\right) q(0) .
\end{array}
$$
... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. (14 points) As shown in Figure 2, the corridor is 3 m wide, the angle between the corridors is 120°, the ground is level, and the ends of the corridor are sufficiently long. Question: What is the maximum length of a horizontal rod (neglecting its thickness) that can pass through the corridor? | II. 9. As shown in Figure 4, draw any horizontal line through the inner vertex $P$ of the corridor corner, intersecting the outer sides of the corridor at points $A$ and $B$. The length of a wooden rod that can pass through the corridor in a horizontal position is less than or equal to $AB$.
Let $\angle B A Q=\alpha$. ... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a>0$, the graphs of the functions $f(x)=|x+2a|$ and $g(x)=|x-a|$ intersect at point $C$, and they intersect the $y$-axis at points $A$ and $B$ respectively. If the area of $\triangle ABC$ is 1, then $a=$ $\qquad$ . | 2. 2 .
From the graphs of $f(x)$ and $g(x)$, we know that $\triangle ABC$ is an isosceles right triangle with base $a$, so its area is $\frac{a^{2}}{4}=1$. Therefore, $a=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the function $y=x^{3}$, the tangent line at $x=a_{k}$ intersects the $x$-axis at point $a_{k+1}$. If $a_{1}=1, S_{n}=\sum_{i=1}^{n} a_{i}$, then $\lim _{n \rightarrow \infty} S_{n}$ $=$ . $\qquad$ | 4.3.
It is known that $y^{\prime}=3 x^{2}$.
Therefore, the equation of the tangent line to $y=x^{3}$ at $x=a_{k}$ is $y-a_{k}^{3}=3 a_{k}^{2}\left(x-a_{k}\right)$.
Thus, the above equation intersects the $x$-axis at the point $\left(a_{k+1}, 0\right)$. Hence, $-a_{k}^{3}=3 a_{k}^{2}\left(a_{k+1}-a_{k}\right)$.
From th... | 3 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
5. The function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies for all $x, y, z \in \mathbf{R}$
$$
f(x+y)+f(y+z)+f(z+x) \geqslant 3 f(x+2 y+z) .
$$
Then $f(1)-f(0)=$ $\qquad$ | 5.0.
Let $x=-y=z$, we get
$$
f(2 x) \geqslant f(0) \Rightarrow f(1) \geqslant f(0) \text {. }
$$
Let $x=y=-z$, we get
$$
f(0) \geqslant f(2 x) \Rightarrow f(0) \geqslant f(1) \text {. }
$$
Thus, $f(1)-f(0)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. (18 points) Let $S$ be a set of distinct quadruples $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$, where $a_{i}=0$ or 1 $(i=1,2,3,4)$. It is known that the number of elements in $S$ does not exceed 15, and satisfies:
if $\left(a_{1}, a_{2}, a_{3}, a_{4}\right) 、\left(b_{1}, b_{2}, b_{3}, b_{4}\right) \in S$, then $\lef... | 12. Clearly, there are 16 possible quadruples. Since at least one quadruple is not in $S$, it follows that $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$, $(0,0,0,1)$ must have at least one that is not in $S$. Otherwise, by the given conditions, all quadruples would be in $S$.
Assume $(1,0,0,0) \notin S$.
In this case, by the g... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Find the units digit of $\left(7^{2004}+36\right)^{818}$.
$(2004$, Shanghai Jiao Tong University Independent Recruitment Examination) | $$
\begin{array}{l}
\text { Solution: Since } 7^{4} \equiv 1(\bmod 10) \\
\Rightarrow 7^{2004} \equiv 1(\bmod 10) \\
\Rightarrow 7^{2004}+36 \equiv 7(\bmod 10) .
\end{array}
$$
$$
\text { Therefore, the original expression } \equiv 7^{818} \equiv 7^{2} \equiv 9(\bmod 10) \text {. }
$$
Thus, the unit digit is 9. | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Let $u$ be a root of the equation
$$
x^{3}-3 x+10=0
$$
Let $f(x)$ be a quadratic polynomial with rational coefficients, and
$$
\alpha=\frac{1}{2}\left(u^{2}+u-2\right), f(\alpha)=u .
$$
Find $f(0)$.
(2010, Five Schools Joint Examination for Independent Enrollment) | From the given, we have $u^{3}-3 u+10=0$. Then
$$
\begin{aligned}
\alpha^{2} &=\frac{1}{4}\left(u^{2}+u-2\right)^{2} \\
&=\frac{1}{4}\left(u^{4}+2 u^{3}-3 u^{2}-4 u+4\right) \\
&=\frac{1}{4}\left[u(3 u-10)+2(3 u-10)-3 u^{2}-4 u+4\right] \\
&=-4-2 u .
\end{aligned}
$$
Let $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{Q}, a \... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Given real numbers $x, y, z$ satisfy
$$
x y z=32, x+y+z=4 \text {. }
$$
Then the minimum value of $|x|+|y|+|z|$ is $\qquad$ .
(2010, Hubei Province High School Mathematics Competition) | Answer: 12.
The text above has been translated into English, maintaining the original text's line breaks and format. Here is the direct output of the translation result. | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given positive integers $a, b$ satisfy
$$
|b-2|+b-2=0,|a-b|+a-b=0,
$$
and $a \neq b$. Then the value of $a b$ is $\qquad$ . | 2.1.2.
From the given conditions, we know that $a>0, b-2 \leqslant 0, a-b \leqslant 0$. Therefore, $a<b \leqslant 2$. Hence, $a=1, b=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 3, in $\triangle ABC$, it is given that $D$ is a point on side $BC$ such that $AD = AC$, and $E$ is the midpoint of side $AD$ such that $\angle BAD = \angle ACE$. If $S_{\triangle BDE} = 1$, then $S_{\triangle ABC}$ is $\qquad$. | 4.4.
Notice that
$$
\begin{array}{l}
\angle E C D=\angle A C D-\angle A C E \\
=\angle A D C-\angle B A D=\angle A B C .
\end{array}
$$
Therefore, $\triangle E C D \backsim \triangle A B C$.
$$
\text { Then } \frac{C D}{B C}=\frac{E D}{A C}=\frac{E D}{A D}=\frac{1}{2} \text {. }
$$
Thus, $D$ is the midpoint of side ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, (20 points) For a certain project, Team A alone needs 12 days to complete, and Team B alone needs 9 days to complete. If the two teams are scheduled to work in whole days, how many schemes are there to ensure that the project is completed within 8 days? | Three, let teams A and B work for $x, y$ days respectively to just meet the requirements. Then we have
$$
\left\{\begin{array}{l}
\frac{x}{12}+\frac{y}{9}=1, \\
0 \leqslant x \leqslant 8, \\
0 \leqslant y \leqslant 8,
\end{array}\right.
$$
and $x, y$ are both integers.
From $\frac{x}{12}+\frac{y}{9}=1$, we get $x=12-y... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the polynomial $f(x)$ satisfy: for any $x \in \mathbf{R}$, we have
$$
f(x+1)+f(x-1)=2 x^{2}-4 x .
$$
Then the minimum value of $f(x)$ is $\qquad$ | - 1. - -2 .
Since $f(x)$ is a polynomial, we know that $f(x+1)$ and $f(x-1)$ have the same degree as $f(x)$. Therefore, $f(x)$ is a quadratic polynomial (let's assume $f(x) = ax^2 + bx + c$). Then,
$$
\begin{array}{l}
f(x+1) + f(x-1) \\
= 2ax^2 + 2bx + 2(a + c) = 2x^2 - 4x.
\end{array}
$$
Thus, $a = 1, b = -2, c = -1... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. If real numbers $x, y$ satisfy the system of equations
$$
\left\{\begin{array}{l}
(x-1)^{2011}+(x-1)^{2009}+2010 x=4020, \\
(y-1)^{2011}+(y-1)^{2009}+2010 y=0,
\end{array}\right.
$$
then $x+y=$ $\qquad$ . | 2. 2 .
Transform the original system of equations into
$$
\begin{array}{l}
\left\{\begin{array}{l}
(x-1)^{2011}+(x-1)^{2009}+2010(x-1)=2010, \\
(y-1)^{2011}+(y-1)^{2009}+2010(y-1)=-2010 .
\end{array}\right. \\
\text { Let } f(t)=t^{2011}+t^{2009}+2010 t(t \in \mathbf{R}) .
\end{array}
$$
Then the function $f(t)$ is a... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $a, b, c \in \mathbf{R}$, and
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c} \text {, }
$$
then there exists an integer $k$, such that the following equations hold for
$\qquad$ number of them.
(1) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k+1}=\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1... | 3. 2 .
From the given equation, we have
$$
\begin{array}{l}
\frac{(b c+a c+a b)(a+b+c)-a b c}{a b c(a+b+c)}=0 . \\
\text { Let } P(a, b, c) \\
=(b c+a c+a b)(a+b+c)-a b c \\
=(a+b)(b+c)(c+a) .
\end{array}
$$
From $P(a, b, c)=0$, we get
$$
a=-b \text { or } b=-c \text { or } c=-a \text {. }
$$
Verification shows that... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given $k_{1}, k_{2}, \cdots, k_{n}$ are $n$ distinct positive integers, and satisfy $\sum_{i=1}^{n} k_{i}^{3}=2024$. Then the maximum value of the positive integer $n$ is $\qquad$ . | 6. 8 .
From the problem, we know that when $n \geqslant 9$, we have
$$
\sum_{i=1}^{n} k_{i}^{3} \geqslant \sum_{i=1}^{n} i^{3} \geqslant \sum_{i=1}^{9} i^{3}=2025>2024,
$$
which is a contradiction. Therefore, $n \leqslant 8$.
Also, $2^{3}+3^{3}+\cdots+9^{3}=2024$, so the maximum value of the positive integer $n$ is 8... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $m, n$ be given positive integers. In each square of an $m \times n$ chessboard, fill a number according to the following rule: first, fill the numbers in the 1st row and the 1st column arbitrarily, then, for any other square, let the number filled in it be $x$, and the number in the 1st row in the same column a... | 6.1.
Let the cell at the $i$-th row and $j$-th column of the chessboard be denoted as $a_{i j}$, and the number filled in cell $a_{i j}$ is also represented by $a_{i j}$.
Consider any rectangle on the chessboard, and let the four corner cells of this rectangle be $a_{i j}, a_{i t}, a_{s j}, a_{s t} (i<s, j<t)$.
By th... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $p$ is a prime number greater than 5, and $m$ is the smallest non-negative remainder when $\left(p^{2}+5 p+5\right)^{2}$ is divided by 120. Then the units digit of $2009^{m}$ is $\qquad$ | Notice that
$$
\begin{array}{l}
\left(p^{2}+5 p+5\right)^{2}=\left[\left(p^{2}+5 p+5\right)^{2}-1\right]+1 \\
=\left(p^{2}+5 p+6\right)\left(p^{2}+5 p+4\right)+1 \\
=(p+1)(p+2)(p+3)(p+4)+1 .
\end{array}
$$
Let $M=(p+1)(p+2)(p+3)(p+4)$.
Clearly, $5! \mid p M$.
Since $p$ is a prime number greater than 5, 120 divides $M$... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $D$, $E$, and $F$ be points on the sides $BC$, $AB$, and $AC$ of $\triangle ABC$, respectively, such that $AE = AF$, $BE = BD$, and $CF = CD$. Given that $AB \cdot AC = 2BD \cdot DC$, $AB = 12$, and $AC = 5$. Then the inradius $r$ of $\triangle ABC$ is $\qquad$ | 4.2.
As shown in Figure 6, let
$$
\begin{array}{l}
A E=A F=x, \\
B E=B D=y, \\
C D=C F=z .
\end{array}
$$
Then $A B=x+y$,
$$
\begin{array}{l}
A C=x+z, \\
B C=y+z .
\end{array}
$$
From $A B \cdot A C=2 B D \cdot D C$
$$
\begin{array}{l}
\Rightarrow(x+y)(x+z)=2 y z \\
\Rightarrow x^{2}+x y+x z=y z \\
\text { Also } A ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given positive real numbers $a$, $b$, $c$ satisfy
$$
(1+a)(1+b)(1+c)=8 \text {. }
$$
Then the minimum value of $a b c+\frac{9}{a b c}$ is | -1.10 .
$$
\begin{array}{l}
\text { Given } 8=(1+a)(1+b)(1+c) \\
\geqslant 2 \sqrt{a} \cdot 2 \sqrt{b} \cdot 2 \sqrt{c}=8 \sqrt{a b c} \\
\Rightarrow a b c \leqslant 1 .
\end{array}
$$
Equality holds if and only if $a=b=c=1$.
It is easy to verify that the function
$$
f(x)=x+\frac{9}{x}
$$
is decreasing on $(0,3)$.
Si... | 10 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. Given a finite arithmetic sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=1$, and a common difference of 2, the arithmetic mean of all its terms is 2011. If one term is removed, the arithmetic mean of the remaining terms is an integer. Then the number of ways to remove a term is $\qquad$. | 4.3.
According to the problem, we have
$$
\frac{1}{n}\left[n+\frac{n(n-1)}{2} \times 2\right]=2011 \text {. }
$$
Solving this, we get $n=2011$.
Thus, the sum of all terms in the sequence is $2011^{2}$.
Suppose that after removing the $k$-th term from the sequence, the arithmetic mean of the remaining terms is an inte... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. As shown in Figure 2, in plane $\alpha$, $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$ are on opposite sides of line $l$, with no common points with $l$, and are symmetric about line $l$. Now, if plane $\alpha$ is folded along line $l$ to form a right dihedral angle, then the six points $A, B, C, A_{1}, B_{1},... | 7.11.
Notice that after the fold, the three sets of four points
$$
\left(A, B, A_{1}, B_{1}\right) 、\left(B, C, B_{1}, C_{1}\right) 、\left(C, A, C_{1}, A_{1}\right)
$$
are all coplanar, therefore, these six points can determine
$$
C_{6}^{3}-3\left(C_{4}^{3}-1\right)=11 \text { (planes). }
$$ | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given points $A, B, C, D$ lie on the same circle, and $BC = DC = 4, AC$ intersects $BD$ at point $E, AE = 6$. If the lengths of segments $BE$ and $DE$ are both integers, find the length of $BD$. | Apply the property to $\triangle B C D$ to get
$$
C E^{2}=C D^{2}-B E \cdot E D=16-6 \cdot E C \text {. }
$$
Solving, we get $E C=2$.
From $B E \cdot E D=A E \cdot E C=12$, and
$$
B D<B C+C D=8 \text {, }
$$
we solve to get $B D=4+3=7$. | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. A circle is filled with 12 positive integers, each taken from $\{1,2, \cdots, 9\}$ (each number can appear multiple times on the circle). Let $S$ represent the sum of all 12 numbers on the circle. If the sum of any three consecutive numbers on the circle is a multiple of 7, then the number of possible values for $S$... | 4.9.
For any three consecutive numbers $a_{k}, a_{k+1}, a_{k+2}$ on a circle, $a_{k}+a_{k+1}+a_{k+2}$ can be 7, 14, or 21.
For any four consecutive numbers on the circle, if they are $a_{k}, a_{k+1}, a_{k+2}, a_{k+3}$, since
$$
a_{k}+a_{k+1}+a_{k+2} \text { and } a_{k+1}+a_{k+2}+a_{k+3}
$$
are both multiples of 7, i... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Let
$$
P=x^{4}+6 x^{3}+11 x^{2}+3 x+31 \text {. }
$$
Find the integer value(s) of $x$ that make $P$ a perfect square. | 10. Since $P=\left(x^{2}+3 x+1\right)^{2}-3(x-10)$, therefore, when $x=10$, $P=131^{2}$ is a perfect square.
Next, we only need to prove: there are no other integer $x$ that satisfy the requirement.
(1) $x>10$.
If $P0$, then $P>\left(x^{2}+3 x\right)^{2}$.
Therefore, $\left(x^{2}+3 x\right)^{2}\left(x^{2}+3 x+1\right)... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. In the sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{1}=2, a_{n+1}-2 a_{n}=2^{n+1}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then the smallest positive integer $n$ for which $a_{n}>10$ holds is $\qquad$ . . | 3.3.
Given $a_{n+1}-2 a_{n}=2^{n+1}$, we know $\frac{a_{n+1}}{2^{n+1}}-\frac{a_{n}}{2^{n}}=1$.
Thus, the sequence $\left\{\frac{a_{n}}{2^{n}}\right\}$ is an arithmetic sequence with a common difference of 1. Also, $a_{1}=2$, so $\frac{a_{n}}{2^{n}}=n \Rightarrow a_{n}=n \times 2^{n}$. Therefore, $\left\{a_{n}\right\}$... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. If the positive integer $m$ makes it true that for any set of positive numbers $a_{1} 、 a_{2} 、 a_{3} 、 a_{4}$ satisfying $a_{1} a_{2} a_{3} a_{4}=1$, we have
$$
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m} \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}
$$
then the minimum value of the positiv... | 9.3.
Let $a_{1}=\frac{1}{27}, a_{2}=a_{3}=a_{4}=3$. Then
$$
\begin{array}{l}
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m}=\left(\frac{1}{27}\right)^{m}+3 \times 3^{m}, \\
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}=27+3 \times \frac{1}{3}=28 .
\end{array}
$$
It is verified that $m=1, m=2$ do not meet t... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
20. Let the ellipse be $\frac{x^{2}}{a^{2}}+y^{2}=1(a>1), \operatorname{Rt} \triangle A B C$ with $A(0,1)$ as the right-angle vertex, and sides $A B, B C$ intersecting the ellipse at points $B, C$. If the maximum area of $\triangle A B C$ is $\frac{27}{8}$, find the value of $a$. | 20. Let $l_{A B}: y=k x+1(k>0)$. Then $l_{A C}: y=-\frac{1}{k} x+1$.
From $\left\{\begin{array}{l}y=k x+1, \\ \frac{x^{2}}{a^{2}}+y^{2}=1,\end{array}\right.$,
$\left(1+a^{2} k^{2}\right) x^{2}+2 a^{2} k x=0$
$\Rightarrow x_{B}=\frac{-2 a^{2} k}{1+a^{2} k^{2}}$.
Thus, $|A B|=\sqrt{1+k^{2}} \cdot \frac{2 a^{2} k}{1+a^{2}... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. If the arithmetic mean of two positive numbers is $2 \sqrt{3}$, and the geometric mean is $\sqrt{3}$, what is the difference between these two numbers? | 1. Let $x, y$ represent two numbers. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
x+y=4 \sqrt{3}, \\
x y=3
\end{array}\right. \\
\Leftrightarrow|x-y|=\sqrt{(x-y)^{2}} \\
=\sqrt{(x+y)^{2}-4 x y}=6,
\end{array}
$$
which means the difference between the two numbers is 6. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. For the quadratic function $y=x^{2}+b x+c$, the vertex of its graph is $D$, and it intersects the positive x-axis at points $A$ and $B$ from left to right, and the positive y-axis at point $C$. If $\triangle A B D$ and $\triangle O B C$ are both isosceles right triangles (where $O$ is the origin), then $b+2 c=$ | 2. 2 .
From the given information, we have
$$
\begin{array}{l}
C(0, c) 、 A\left(\frac{-\dot{b}-\sqrt{b^{2}-4 c}}{2}, 0\right), \\
B\left(\frac{-b+\sqrt{b^{2}-4 c}}{2}, 0\right), D\left(-\frac{b}{2},-\frac{b^{2}-4 c}{4}\right) .
\end{array}
$$
Draw $D E \perp A B$ at point $E$. Then $2 D E=A B$,
which means $2 \times ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The value of the positive integer $n$ that makes $2^{n}+256$ a perfect square is $\qquad$ | 3. 11.
When $n8$, $2^{n}+256=2^{8}\left(2^{n-8}+1\right)$, if it is a perfect square, then $2^{n-8}+1$ is the square of an odd number.
Let $2^{n-8}+1=(2 k+1)^{2}$ ( $k$ is a natural number). Then $2^{n-10}=k(k+1)$.
Since $k$ and $k+1$ are one odd and one even, then $k=1$, thus, $n=11$. | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 When $1 \leqslant x \leqslant 2$, simplify
$$
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}
$$
The value equals $\qquad$ [5]
(2009, Beijing Middle School Mathematics Competition (Grade 8)) | Solution 1
$$
p=\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}} \text {. }
$$
Then $p^{2}=2 x+2 \sqrt{x^{2}-(2 \sqrt{x-1})^{2}}$
$$
=2 x+2 \sqrt{(x-2)^{2}} \text {. }
$$
Since $1 \leqslant x \leqslant 2$, therefore,
$$
\begin{array}{l}
p^{2}=2 x+2 \sqrt{(x-2)^{2}} \\
=2 x+2(2-x)=4 .
\end{array}
$$
And $p \geqslant 0$, s... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $f(x)$ be a continuous even function, and when $x>0$, $f(x)$ is a strictly monotonic function. Then the sum of all $x$ that satisfy $f(x) = f\left(\frac{x+3}{x+4}\right)$ is ( ).
(A) -3
(B) -8
(C) 3
(D) 8 | 8. B.
When $f(x)=f\left(\frac{x+3}{x+4}\right)$, i.e., $x=\frac{x+3}{x+4}$, we get $x^{2}+3 x-3=0$. At this point,
$$
x_{1}+x_{2}=-3 \text {. }
$$
Since $f(x)$ is a continuous even function, another scenario is $f(-x)=f\left(\frac{x+3}{x+4}\right)$, i.e., $-x=\frac{x+3}{x+4}$, which gives
$$
x^{2}+5 x+3=0 \text {. }
... | -8 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 6 Let $a=\frac{\sqrt{5}-1}{2}$. Then $\frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a}=$ $\qquad$ [6] $(2008$, National Junior High School Mathematics Competition) | Solve: Given $a=\frac{\sqrt{5}-1}{2} \Rightarrow 2 a+1=\sqrt{5}$.
Square both sides and simplify to get
$$
a^{2}+a=1 \text {. }
$$
Therefore, the original expression is
$$
\begin{array}{l}
=\frac{a^{3}\left(a^{2}+a\right)-2 a^{3}-\left(a^{2}+a\right)+2}{a(a+1)(a-1)} \\
=\frac{1-a^{3}}{\left(a^{2}+a\right)(a-1)} \\
=-\... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given $a=\sqrt{5}-1$. Then the value of $2 a^{3}+7 a^{2}-2 a$ -12 is $\qquad$ [7]
(2010) "Mathematics Weekly" Cup National Junior High School Mathematics Competition) | Given $a=\sqrt{5}-1 \Rightarrow a+1=\sqrt{5}$. Squaring both sides, we get
$$
a^{2}+2 a=4 \text {. }
$$
Therefore, the original expression is
$$
\begin{array}{l}
=2 a\left(a^{2}+2 a\right)+3 a^{2}-2 a-12 \\
=3 a^{2}+6 a-12 \\
=3\left(a^{2}+2 a\right)-12=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Let $M=(5+\sqrt{24})^{2 n}\left(n \in \mathbf{N}_{+}\right), N$ be the fractional part of $M$. Then the value of $M(1-N)$ is $\qquad$ | 10.1.
Since $(5+\sqrt{24})^{2 n}+(5-\sqrt{24})^{2 n}$ is a positive integer, and $0<(5-\sqrt{24})^{2 n}<1$, therefore,
$$
\begin{array}{l}
N=1-(5-\sqrt{24})^{2 n} \\
M(1-N)=(5+\sqrt{24})^{2 n}(5-\sqrt{24})^{2 n}=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
16. As shown in Figure $2, P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and the circle $(x-1)^{2}+y^{2}=1$ is inscribed in $\triangle P B C$. Find the minimum value of the area of $\triangle P B C$. | 16. Let \( P\left(x_{0}, y_{0}\right) \), \( B(0, b) \), \( C(0, c) \), and assume \( b > c \). The line \( l_{P B} \) is given by \( y - b = \frac{y_{0} - b}{x_{0}} x \), which can be rewritten as
\[
\left(y_{0} - b\right) x - x_{0} y + x_{0} b = 0.
\]
The distance from the circle's center \((1, 0)\) to \( P B \) is ... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $x$ be a positive integer, and $x<50$. Then the number of $x$ such that $x^{3}+11$ is divisible by 12 is $\qquad$. | $$
\begin{array}{l}
x^{3}+11=(x-1)\left(x^{2}+x+1\right)+12 \\
=(x-1) x(x+1)+(x-1)+12 .
\end{array}
$$
Notice that $6 \mid(x-1) x(x+1)$, so $6 \mid(x-1)$;
also, $2 \mid \left(x^{2}+x+1\right)$, hence $12 \mid (x-1)$.
Since $x<50$, we have
$$
x-1=0,12,24,36,48 \text {. }
$$
Thus, $x=1,13,25,37,49$, a total of 5. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 5, in $\triangle A B C$, $D$ is the midpoint of side $A B$, and $G$ is the midpoint of $C D$. A line through $G$ intersects $A C$ and $B C$ at points $P$ and $Q$, respectively. Then the value of $\frac{C A}{C P}+\frac{C B}{C Q}$ is $\qquad$. | 4. 4 .
As shown in Figure 12, draw $A E / / P Q$ intersecting $C D$ at point $E$, and $B F / / P Q$ intersecting the extension of $C D$ at point $F$. Then $A E / / B F$.
Since $A D = B D$, therefore, $E D = F D$.
$$
\begin{array}{c}
\text { Also, } \frac{C A}{C P} = \frac{C E}{C G}, \\
\frac{C B}{C O} = \frac{C F}{C ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. If $n \in \mathbf{N}, n \geqslant 2, a_{i} \in\{0,1, \cdots, 9\}$
$$
\begin{array}{l}
(i=1,2, \cdots, n), a_{1} a_{2} \neq 0 \text {, and } \\
\sqrt{a_{1} a_{2} \cdots a_{n}}-\sqrt{a_{2} a_{3} \cdots a_{n}}=a_{1}, \\
\end{array}
$$
then $n=$ $\qquad$, where $\overline{a_{1} a_{2} \cdots a_{n}}$ is the $n$-digit num... | 2. 2 .
Let $x=\overline{a_{2} a_{3} \cdots a_{n}} \in \mathbf{N}_{+}$. Then $\overline{a_{1} a_{2} \cdots a_{n}}=10^{n-1} a_{1}+x$.
By the given condition, $\sqrt{10^{n-1} a_{1}+x}-\sqrt{x}=a_{1}$. Therefore, $10^{n-1} a_{1}+x=x+a_{1}^{2}+2 a_{1} \sqrt{x}$.
Thus, $10^{n-1}-2 \sqrt{x}=a_{1}$.
If $n \geqslant 3$, then
$... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. On a straight line, three points $A$, $B$, and $C$ are arranged in sequence, and $A B=6, A C=24, D$ is a point outside the line, and $D A$ $\perp A B$. When $\angle B D C$ takes the maximum value, $A D=$ $\qquad$ . | 5. 12 .
Let $\angle B D C=\theta\left(\theta<90^{\circ}\right), \triangle B C D$'s circumcircle $\odot O$ has a radius of $R$. Then $\sin \theta=\frac{B C}{2 R}$.
When $R$ decreases, $\theta$ increases.
Therefore, when $\odot O$ is tangent to $A D$ at point $D$, $\theta$ is maximized.
At this time, $A D^{2}=A B \cdot ... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the set $M=\{2,0,11\}$. If $A \varsubsetneqq M$, and $A$ contains at least one even number, then the number of sets $A$ that satisfy the condition is $\qquad$ . | 1. 5.
The sets $A$ that satisfy the conditions are $\{2\}, \{0\}, \{2,0\}, \{2,11\}, \{0,11\}$. There are 5 in total. | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Let the function
$$
f(x)=\left\{\begin{array}{ll}
\frac{1}{p}, & x=\frac{q}{p} ; \\
0, & x \neq \frac{q}{p},
\end{array}\right.
$$
where $p$ and $q$ are coprime, and $p \geqslant 2$. Then the number of $x$ values that satisfy $x \in[0,1]$ and $f(x)>\frac{1}{5}$ is $\qquad$ . | 7.5.
Obviously, $x=\frac{q}{p}$ (otherwise, $f(x)=0$). At this point, from $f(x)=\frac{1}{p}>\frac{1}{5}$, we get $p<5$, i.e., $p=2,3,4$. When $p=2$, $x=\frac{1}{2}$; when $p=3$, $x=\frac{1}{3}, \frac{2}{3}$; when $p=4$, $x=\frac{1}{4}, \frac{3}{4}$.
Therefore, there are 5 values of $x$ that satisfy the condition. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Find all odd prime numbers $p$ such that
$$
p \mid \sum_{k=1}^{2011} k^{p-1} .
$$ | If $p>2011$, then for $k(1 \leqslant k \leqslant 2011)$ by Fermat's Little Theorem we have
$k^{p-1} \equiv 1(\bmod p)$.
Thus $\sum_{k=1}^{2011} k^{p-1} \equiv 2011 \not\equiv 0(\bmod p)$, a contradiction.
Therefore, $p \leqslant 2011$.
Let $2011=p q+r(0 \leqslant r < q)$, then if $r>q$, we have $q=r$, so
$$
2011=p q+r=... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. The equation concerning $x, y$
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2011}
$$
has $\qquad$ groups of positive integer solutions $(x, y)$. | 3. 12 .
From $\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2011}$, we get
$$
\begin{array}{l}
x y-2011 x-2011 y-2011=0 \\
\Rightarrow(x-2011)(y-2011) \\
\quad=2011 \times 2012 \\
=2^{2} \times 503 \times 2011 .
\end{array}
$$
Thus, the positive integer solutions of the original equation are
$$
(2+1)(1+1)(1+1)=12 \t... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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